Question: $\begin{cases}d(1)=3\\\\ d(n)=d(n-1)-14 \end{cases}$ Find the $3^{\text{rd}}$ term in the sequence.
Explanation: This is a recursive formula. It tells us that the first term is $3$ and that the common difference is $-14$. $\begin{aligned} {d(1)}&=3 \\\\ {d(2)}&={d(1)}-14=-11 \\\\ {d(3)}&={d(2)}-14=-25 \end{aligned}$ The $3^{\text{rd}}$ term is $-25$.